Grushko’s Lemma. Suppose is surjective and is minimal. If , then such that for .

Pf. Let be simplicial and let be a graph of spaces with vertex spaces and edge space a point. So where .

Let be a graph so that and realize as a simplicial map . Let . Because is minimal, is a forest, contained in . The goal is to modify by a homotopy to reduce the number of connected components of .

Let be the component that contains . Let be some other component. Let a path in from to .

Look at . Because is surjective, there exists such that . Therefore if , then is null-homotopic in and gives a path from to .

We can write as a concaternation as such that for each , . By the Normal Form Theorem, there exists such that is null-homotopic in .

We can now modify by a homotopy so that . Therefore and the number of components of has gone down. By induction, we can choose so that is a tree. Now factors through . Then and there is a unique vertex of that maps to . So every simple loop in is either contained in or as required.

An immediate consequence is that .

Grushko’s Theorem. Let be finitely generated. Then where each is freely indecomposable and is free. Furthermore, the integers and are unique and the are unique up to conjugation and reordering.

Pf. Existence is an immediate corollary of the fact that rank is additive.

Suppose . Let be the graph of groups. Let be the Bass-Serre tree of .

Consider the action of on . Because is freely indecomposable, stabilize a vertex of . Therefore is conjugate into some .

Now consider the action of on . is a graph of groups with underlying graph , say, and is a free factor in . But there is a covering map that induces a surjection . Therefore, . The other inequality can be obtained by switching and .

Last time, we used the following lemma without justification, so let’s prove it now.

Lemma 30. Let be a graph of groups with finite and finitely generated. If is finitely generated for every edge , then is finitely generated for every .

This is not completely trivial: it is certainly possible for finitely generated group to have subgroups that are not finitely generated. Indeed, recall the proof that every countable group embeds in the free group on three generators.

Pf. Let be a finite generating set for , and for each let be a finite generating set for the edge group . By the Normal Form Theorem, every can be written in the form

where each is a stable letter and each for some . For a fixed , let

,

where for each adjoining the plus or minus is chosen so that . It is clear then that is contained in . To see that is finite, note that since is finite, the first union is finite; and since is finite there can be only finitely many edges adjoining a given vertex, so the second union is finite. Hence it remains to prove that generates .

Let . Because generates , we have

where . Each has a normal form as above, so we get an expression of the form

, or .

By the Normal Form Theorem, the expression on the right can then be simplified. After the simplification process, we have no stable letters left, and every is either contained in or is a product of elements of the incident edge groups, and in both cases lie in .

Remember that Theorem 19 said is LERF, answering our question (b). But in fact, we get more from the proof of Theorem 19.

Definition. Recall that is a retract if the inclusion has a left inverse . Similarly, we call a virtual retract if is a retract of a finite index subgroup of .

For instance, Marshall Hall’s Theorem implies that every finitely generated subgroup of a free group is a virtual retract.

Theorem 20. Every finitely generated subgroup of is a virtual retract.

Pf. Consider the setup of the proof of Theorem 19. We start with a subgroup and end up with a finitely sheeted covering space . The graph of spaces is built using the “obvious” bijection between elevations to and elevations to . Thus the identification extends to a topological retraction . Now, we shall build a map . We build it at each vertex space, one at a time. From the proof of Lemma 29, we see that the core of each is a topological retract of the corresponding . Furthermore, we can choose the retraction so that for each long loop of degree that we added is mapped to a null-homotopic loop in . This allows you to piece together the map into a retraction .

Exercise 14 asserted that a (virtual) retract is quasi-isometrically embedded, and so Theorem 20 has the following corollary:

Lemma 29: Suppose is a finite set of infinite degree elvations and is compact. Then for all sufficiently large , there exists an intermediate covering such that

(a) embeds in

(b) every descends to an elevation of degree

(c) the are pairwise distinct

Proof: We claim that the images of never share an infinite ray (a ray is an isometric embedding of ). Neither do two ends of the same elevation. Let’s claim by passing to the universal cover of, a tree .

For each , lift to a map . If and share an infinite ray then there exists such that and overlay in an infinite ray. The point is that correspond to cosets and . But this implies that

This implies that . So . A similar argument implies that the two ends of do not overlap in an infinite ray. This proves the claim.

Let be the core of . Enlarging if necessary, we can assume that

(i) ;

(ii) is a connected subgraph;

(iii) for each , for some , ;

(iv) for each , .

For each identifying with so that is identified with and is identified with . Let

For all sufficiently large ,

Now, the restriction of factors through , where . This is a finite-to-one immersion, so, by theorem 5, we can complete it to a finite-sheeted covering map as required.

Theorem 19: is LERF.

Recall the set-up from the previous lecture. We built a graph of spaces for .

Proof: Let be finitely generated. Let be the corresponding covering space of and let be compact. Because is finitely generated, there exists a subgraph of spaces such that . We can take large enough so that . We can enlarge so that it contains every finite-degree edge space of . Also enlarge so that

for any . For each let and let incident edge map of infinite degree .

Applying lemma 29 to , for some large , set . (Here we use the fact that vertex groups of are finitely generated)

Define as follows:

For each , the edge space is the that the lemma produced from the corresponding .

Now, by construction, can be completed to a graph of spaces so that the map

factors through and embeds. Let be identical to except with +’s and -‘s exchanged. Clearly satisfies Stallings condition, as required.